Question: You have found the following ages (in years) of all 5 gorillas at your local zoo: $ 11,\enspace 5,\enspace 2,\enspace 2,\enspace 12$ What is the average age of the gorillas at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 5 gorillas at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{11 + 5 + 2 + 2 + 12}{{5}} = {6.4\text{ years old}} $ Find the squared deviations from the mean for each gorilla. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $11$ years $4.6$ years $21.16$ years $^2$ $5$ years $-1.4$ years $1.96$ years $^2$ $2$ years $-4.4$ years $19.36$ years $^2$ $2$ years $-4.4$ years $19.36$ years $^2$ $12$ years $5.6$ years $31.36$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{21.16} + {1.96} + {19.36} + {19.36} + {31.36}} {{5}} $ $ {\sigma^2} = \dfrac{{93.2}}{{5}} = {18.64\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{18.64\text{ years}^2}} = {4.3\text{ years}} $ The average gorilla at the zoo is 6.4 years old. There is a standard deviation of 4.3 years.